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Compute the temperature of point P(2.5, 2.5) using natural coordinate system for quadrilateral element shown fig

enter image description here

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enter image description here

x, y = 2.5, 2.5

x1,y1 = 2, 2

x2,y2 = 4, 2

x3,y4 = 4, 3

x4,y4 = 2, 4

Converting into-local coordinate,

enter image description here

Now, ϕ1=14(1ξ)(1η)

ϕ2=14(1+ξ)(1η)

ϕ3=14(1+ξ)(1η)

ϕ4=14(1+ξ)(1η)

Now, ˉx=ϕjˉxj

ˉx = ϕ1ˉx1+ϕ1ˉx2+ϕ1ˉx3+ϕ1ˉx4

0.5 = 14(1ξ)(1η)ˉx1+14(1+ξ)(1η)ˉx2+14(1+ξ)(1+η)ˉx3+14(1ξ)(1+η)ˉx1

i.e. 12=14(1+ξ)(1η)2+14(1ξ)(1+η)2

i.e 12=12(1+ξ)(1η)+12+(1+ξ)(1+η)

1=1+ξ+ηηξ+1+ξ+η+ηξ

1=2+2ξ

ξ12

= (1 + ξ)(1 − η) + (1 + ξ)(1 + η)

Now, ˉy=ϕjˉyj

0.5=140.5=(1ξ)(1η)ˉy1+14(1+ξ)(1η)barx2+14(1+ξ)(1+η)ˉx3+14(1ξ)(1+η)ˉx4

0.5 =14(1+ξ)(1+η)+14(1ξ)(1+η)

12=14[1+ξ+η+ηξ+1ξ+ηηξ

∴ 2 = 2 +2η=>2η=0=>η=0

Therefore, point P is given by (-1/2,0) in ξ,ηcoordinate,

To compute the result,

ϕ1=14(1ξ)(1η)=14(1+12)1=38

ϕ2=14(1ξ)(1η)=14(1+12)1=18

ϕ3=14(1ξ)(1η)=14(1+12)1=18

ϕ4=14(1ξ)(1η)=14(1+12)1=38

Now,

ϕ1+ϕ2+ϕ3+ϕ4=38+18+18+38=88

ϕ1+ϕ2+ϕ3+ϕ4

Now, Temperature at point P,

TP=ϕ1T1+ϕ2T2+ϕ3T3+ϕ4T4

T1=100CT2=60CT3=50CT4=90C

TP=38100+1860+1850+3890

TP=85C

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